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        <p>最近看到很多这种lcs与lis得题目，就来写一个关于这种题目的一个总结<br>更新<br>经我仔细思考，之前有以下漏洞：</p>
<ol>
<li>二分法$lis$已经把树状数组所有功能全部重合，具体写法写到里面了。<a id="more"></a>
<h1 id="最长上升子序列"><a href="#最长上升子序列" class="headerlink" title="最长上升子序列"></a>最长上升子序列</h1>顾名思义，就是求最长上升子序列的最大长度，这样的求法一共有三种！<br>分别介绍一下：<h2 id="n-2-的-dp-写法"><a href="#n-2-的-dp-写法" class="headerlink" title="$n^2$的$dp$写法"></a>$n^2$的$dp$写法</h2>定义$dp[i]$为从第$i$个数字结尾的上升子序列(该上升子序列一定以$a[i]$结尾)的长度。我们每一次枚举一个$j$，往前枚举$i$找到一个比$a[j]$小的最大的$dp[i]$，更新$dp[j]$。<br>换句话说就是找到前面一个比当前值小且以他结尾的最长上升子序列。<br>代码</li>
</ol>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line">int a[MAXN], d[MAXN];</span><br><span class="line">int dp() &#123;</span><br><span class="line">  d[1] = 1;</span><br><span class="line">  int ans = 1;</span><br><span class="line">  for (int i = 2; i &lt;= n; i++) &#123;</span><br><span class="line">    for (int j = 1; j &lt; i; j++)</span><br><span class="line">      if (a[j] &lt; a[i]) &#123;</span><br><span class="line">        d[i] = max(d[i], d[j] + 1);</span><br><span class="line">        ans = max(ans, d[i]);</span><br><span class="line">      &#125;</span><br><span class="line">  &#125;</span><br><span class="line">  return ans;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="二分-贪心法"><a href="#二分-贪心法" class="headerlink" title="二分+贪心法"></a>二分+贪心法</h2><h3 id="思想"><a href="#思想" class="headerlink" title="思想"></a>思想</h3><p>对于一个序列，我们枚举一个值，考虑这个值如果在最长$LIS$里面，那么这个值的位置所在的地方是在哪？在前面比该值小的后面。也就是说，这样对于每一个数字，我们都可以知道，如果他在最长上升序列的位置，记录一下最大的位置就是最长长度！</p>
<h3 id="实现"><a href="#实现" class="headerlink" title="实现"></a>实现</h3><p>拿一个数组存储在最长上升子序列中第$i$个位置所能出现的最小值，后面不用更新。</p>
<h3 id="举例"><a href="#举例" class="headerlink" title="举例"></a>举例</h3><blockquote>
<p>2 5 1 3 4<br>2<br>2 5<br>1 5//更新第一个位置的值，使其变小<br>1 3//更新第二个位置的值，使其变小<br>1 3 4</p>
</blockquote>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line">for (int i = 0; i &lt; n; ++i) scanf(&quot;%d&quot;, a + i);</span><br><span class="line">memset(dp, 0x1f, sizeof dp);//预处理为最大值</span><br><span class="line">mx = dp[0];</span><br><span class="line">for (int i = 0; i &lt; n; ++i) &#123;</span><br><span class="line">  pos = lower_bound(dp, dp + n, a[i]) - dp;//找到大于等于a[i]的第一个值，然后把a[pos]变小。</span><br><span class="line">  dp[pos] = a[i];//注意此时的dp[1~pos]就是以a[i]结尾的最长上升子序列，树状数组。。。多余了</span><br><span class="line">&#125;</span><br><span class="line">ans = 0;</span><br><span class="line">while (dp[ans] != mx) ++ans;</span><br></pre></td></tr></table></figure>
<p>复杂度$O(Nlog(N))$，该方法也可以$nlogn$求具体最大字串，具体实现可自己想。(提示：参照前向星写法)</p>
<h2 id="树状数组写法"><a href="#树状数组写法" class="headerlink" title="树状数组写法"></a>树状数组写法</h2><p><strong><a href="https://oi-wiki.org/ds/fenwick/" target="_blank" rel="noopener">不了解树状数组请参考</a></strong><br>可能好奇上面已经够了呀，干嘛还要这个写法呢？给出一点，他能求出具体的以a[i]结尾的最长上升子序列字符串。(二分法也行，不过已经写了，就不删除了，当思维扩展吧)</p>
<h3 id="思想-1"><a href="#思想-1" class="headerlink" title="思想"></a>思想</h3><p>树状数组可以连续修改一个区间，以及查询一个区间。<br>我们从第一个值开始枚举，每次按照贪心思想，把比他大的值全部更新为当前值！那么就可以得到以$a[i]$结尾的最大上升子序列的具体值了。</p>
<h3 id="实现-1"><a href="#实现-1" class="headerlink" title="实现"></a>实现</h3><p>首先离散化，<a href="https://oi-wiki.org/misc/discrete/" target="_blank" rel="noopener">不懂点击</a>，然后用树状数组查出比第$i$个值大的第一个值得位置，然后对该位置以及以后位置更新为当前值！</p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br></pre></td><td class="code"><pre><span class="line">#include&lt;bits/stdc++.h&gt;</span><br><span class="line">#define REP(i, a, b) for(register int i = (a); i &lt; (b); i++)</span><br><span class="line">#define _for(i, a, b) for(register int i = (a); i &lt;= (b); i++)</span><br><span class="line">using namespace std;</span><br><span class="line"></span><br><span class="line">const int MAXN = 1e3 + 10;</span><br><span class="line">int a[MAXN], b[MAXN], n, m, ans; </span><br><span class="line">int dp[MAXN], f[MAXN];</span><br><span class="line"></span><br><span class="line">inline int lowbit(int x) &#123; return x &amp; (-x); &#125; </span><br><span class="line"></span><br><span class="line">void motify(int x, int p)</span><br><span class="line">&#123;</span><br><span class="line">    for(; x &lt;= m; x += lowbit(x))</span><br><span class="line">        f[x] = max(f[x], p);</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int get_max(int x)</span><br><span class="line">&#123;</span><br><span class="line">    int res = 0;</span><br><span class="line">    for(; x; x -= lowbit(x))</span><br><span class="line">        res = max(res, f[x]);</span><br><span class="line">    return res;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main()</span><br><span class="line">&#123;</span><br><span class="line">    scanf(&quot;%d&quot;, &amp;n);</span><br><span class="line">    _for(i, 1, n) scanf(&quot;%d&quot;, &amp;a[i]), b[i] = a[i];</span><br><span class="line">    sort(b + 1, b + n + 1);</span><br><span class="line">    m = unique(b + 1, b + n + 1) - b - 1;</span><br><span class="line">    _for(i, 1, n) a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;</span><br><span class="line">    int ans = 0;</span><br><span class="line">    _for(i, 1, n)</span><br><span class="line">    &#123;</span><br><span class="line">        dp[i] = get_max(a[i]) + 1;//查询大于等于a[i]得第一个位置</span><br><span class="line">        ans = max(ans, dp[i]);//最长上升子序列</span><br><span class="line">        motify(a[i], dp[i]); //把该位置往后得所有值全部更新</span><br><span class="line">    &#125;</span><br><span class="line">    printf(&quot;%d\n&quot;, ans);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>这个大家可能不熟悉，写全了。(百度抄的，改成了这样子的)</p>
<h1 id="最长公共子序列"><a href="#最长公共子序列" class="headerlink" title="最长公共子序列"></a>最长公共子序列</h1><p>这里有两种解决办法</p>
<h2 id="动态规划-n-2"><a href="#动态规划-n-2" class="headerlink" title="动态规划($n^2$)"></a>动态规划($n^2$)</h2><p>$dp[i][j]$表示从字符串a的前i个字符与字符串b的前j个字符串的最长公共子序列的长度。说到这里大家伙可能就都知道了，直接贴代码，代码里面解释。<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line">for(i=1;i&lt;=n+1;i++)&#123;</span><br><span class="line">    for(j=1;j&lt;=m+1;j++)&#123;</span><br><span class="line">        if(a[i]==b[j]) dp[i][j]=dp[i-1][j-1]+1;//俩子序列结尾相同时为前面值加一</span><br><span class="line">        else dp[i][j]=max(dp[i][j-1],dp[i-1][j]);//不同时为这个。</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure><br>这个路径可以利用转移方程按前向星思路保存路径</p>
<h2 id="LCS-转-LIS"><a href="#LCS-转-LIS" class="headerlink" title="$LCS$转$LIS$"></a>$LCS$转$LIS$</h2><p><strong>使用条件，字串里面元素不得出现重复</strong></p>
<h3 id="思想-2"><a href="#思想-2" class="headerlink" title="思想"></a>思想</h3><p>我们按照第一个字串的顺序映射第一个字串元素在第二个子串中的位置(数组C)。那么$LCS$就转为$LIS$了。没看懂的话，我下面解释一下：<br>第一个字串按顺序映射位置，那么当位置单调递增，也就是在第二个字串中按顺序排列的，那么这几个位置就会是最长公共子序列！</p>
<h3 id="举例-1"><a href="#举例-1" class="headerlink" title="举例"></a>举例</h3><p>假设两个字串</p>
<blockquote>
<p>a 1 7 5 4 8 3 9<br>b 1 4 3 5 6 2 8 9</p>
</blockquote>
<p>映射后</p>
<blockquote>
<p>a 1 7 5 4 8 3 9<br>b 1 4 3 5 6 2 8 9<br>c 1 0 4 2 0 0 7 8</p>
</blockquote>
<p>C数组$LIS$得到答案和路径长度4，路径1 4 7 8<br>到B里面就是 $1b[1] 5(b[4]) 8(b[7]) 9(b[8])$</p>
<h3 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a>代码</h3><p>下面是二分法，树状数组请自己改编<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line">for(i=1;i&lt;=n+1;i++) vis[a[i]]=i;</span><br><span class="line">for(i=1;i&lt;=m+1;i++) b[i]=vis[b[i]];</span><br><span class="line">for(i=1;i&lt;=m+1;i++) dp[i]=2e9;</span><br><span class="line">for(i=1;i&lt;=m+1;i++)&#123;</span><br><span class="line">    tot = lower_bound(dp+1,dp+m+1,b[i])-dp;</span><br><span class="line">    ans=max(ans,tot);</span><br><span class="line">    dp[tot]=b[i];</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>
<h1 id="练习题目"><a href="#练习题目" class="headerlink" title="练习题目"></a>练习题目</h1><p><a href="https://www.luogu.com.cn/problem/P1020" target="_blank" rel="noopener">导弹拦截</a><br><a href="https://vjudge.net/problem/UVA-10635" target="_blank" rel="noopener">LCS</a></p>
<h1 id="参考资料"><a href="#参考资料" class="headerlink" title="参考资料"></a>参考资料</h1><p><a href="https://www.cnblogs.com/sugewud/p/9823222.html" target="_blank" rel="noopener">树状数组得最长不下降子序列</a><br><a href="https://oi-wiki.org/ds/fenwick/" target="_blank" rel="noopener">树状数组</a><br><a href="https://oi-wiki.org/misc/discrete/" target="_blank" rel="noopener">离散话</a><br><a href="https://oi-wiki.org/dp/basic/" target="_blank" rel="noopener">oi-wiki</a></p>

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